Question

A radioactive counter gave a reading of 425 min^-1 for a 13.9 mg sample of cobalt(II) chloride partially enriched with cobalt-60 (t_1/2 = 5.2 yr). What percentage of the cobalt atoms is cobalt-60?

percent

The number of significant digits is set to 3; the tolerance is +/-4%

Answer 1

The objective is to calculate the percentage of the cobalt atoms to the following data. Ans Given that, 425 425 =7.08 5. Rate, r= I min 605 Half-life , t1/2 = 5.2 yn = 5.2 you = 1.64 x 10°S. -9 = 4.23X10 Rate constant , ka In 2 ti In 2 1.64x108s VE KN. NE Number of radioactive nuclei to N. Rearrange r Na K substitute 7.085) Na 4.23x10951 N=16738 x 10 9 nuclei use the below expression to find the number of nuclei in 13.9 mg. 1g imol X 13.9 mg x 103 mg * 129.8399 6.023x10 nuclei i mol 19 No=6.4480 x 10 nuclei

Percentage * 100% No substitute percentage 1.6738x10? nuclei X 100%. 6.4480 x 10' nuclei Percentage = -9 2.596X10'y